\(\int x^3 (a+b \log (c x^n)) \log (d (e+f x^2)^m) \, dx\) [96]

Optimal result

Integrand size = 26, antiderivative size = 221 \[ \int x^3 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right ) \, dx=-\frac {3 b e m n x^2}{16 f}+\frac {1}{16} b m n x^4+\frac {e m x^2 \left (a+b \log \left (c x^n\right )\right )}{4 f}-\frac {1}{8} m x^4 \left (a+b \log \left (c x^n\right )\right )+\frac {b e^2 m n \log \left (e+f x^2\right )}{16 f^2}+\frac {b e^2 m n \log \left (-\frac {f x^2}{e}\right ) \log \left (e+f x^2\right )}{8 f^2}-\frac {e^2 m \left (a+b \log \left (c x^n\right )\right ) \log \left (e+f x^2\right )}{4 f^2}-\frac {1}{16} b n x^4 \log \left (d \left (e+f x^2\right )^m\right )+\frac {1}{4} x^4 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )+\frac {b e^2 m n \operatorname {PolyLog}\left (2,1+\frac {f x^2}{e}\right )}{8 f^2} \] Output:

-3/16*b*e*m*n*x^2/f+1/16*b*m*n*x^4+1/4*e*m*x^2*(a+b*ln(c*x^n))/f-1/8*m*x^4 
*(a+b*ln(c*x^n))+1/16*b*e^2*m*n*ln(f*x^2+e)/f^2+1/8*b*e^2*m*n*ln(-f*x^2/e) 
*ln(f*x^2+e)/f^2-1/4*e^2*m*(a+b*ln(c*x^n))*ln(f*x^2+e)/f^2-1/16*b*n*x^4*ln 
(d*(f*x^2+e)^m)+1/4*x^4*(a+b*ln(c*x^n))*ln(d*(f*x^2+e)^m)+1/8*b*e^2*m*n*po 
lylog(2,1+f*x^2/e)/f^2
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.24 (sec) , antiderivative size = 324, normalized size of antiderivative = 1.47 \[ \int x^3 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right ) \, dx=-\frac {-4 a e f m x^2+3 b e f m n x^2+2 a f^2 m x^4-b f^2 m n x^4-4 b e f m x^2 \log \left (c x^n\right )+2 b f^2 m x^4 \log \left (c x^n\right )+4 b e^2 m n \log (x) \log \left (1-\frac {i \sqrt {f} x}{\sqrt {e}}\right )+4 b e^2 m n \log (x) \log \left (1+\frac {i \sqrt {f} x}{\sqrt {e}}\right )+4 a e^2 m \log \left (e+f x^2\right )-b e^2 m n \log \left (e+f x^2\right )-4 b e^2 m n \log (x) \log \left (e+f x^2\right )+4 b e^2 m \log \left (c x^n\right ) \log \left (e+f x^2\right )-4 a f^2 x^4 \log \left (d \left (e+f x^2\right )^m\right )+b f^2 n x^4 \log \left (d \left (e+f x^2\right )^m\right )-4 b f^2 x^4 \log \left (c x^n\right ) \log \left (d \left (e+f x^2\right )^m\right )+4 b e^2 m n \operatorname {PolyLog}\left (2,-\frac {i \sqrt {f} x}{\sqrt {e}}\right )+4 b e^2 m n \operatorname {PolyLog}\left (2,\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{16 f^2} \] Input:

Integrate[x^3*(a + b*Log[c*x^n])*Log[d*(e + f*x^2)^m],x]
 

Output:

-1/16*(-4*a*e*f*m*x^2 + 3*b*e*f*m*n*x^2 + 2*a*f^2*m*x^4 - b*f^2*m*n*x^4 - 
4*b*e*f*m*x^2*Log[c*x^n] + 2*b*f^2*m*x^4*Log[c*x^n] + 4*b*e^2*m*n*Log[x]*L 
og[1 - (I*Sqrt[f]*x)/Sqrt[e]] + 4*b*e^2*m*n*Log[x]*Log[1 + (I*Sqrt[f]*x)/S 
qrt[e]] + 4*a*e^2*m*Log[e + f*x^2] - b*e^2*m*n*Log[e + f*x^2] - 4*b*e^2*m* 
n*Log[x]*Log[e + f*x^2] + 4*b*e^2*m*Log[c*x^n]*Log[e + f*x^2] - 4*a*f^2*x^ 
4*Log[d*(e + f*x^2)^m] + b*f^2*n*x^4*Log[d*(e + f*x^2)^m] - 4*b*f^2*x^4*Lo 
g[c*x^n]*Log[d*(e + f*x^2)^m] + 4*b*e^2*m*n*PolyLog[2, ((-I)*Sqrt[f]*x)/Sq 
rt[e]] + 4*b*e^2*m*n*PolyLog[2, (I*Sqrt[f]*x)/Sqrt[e]])/f^2
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 214, normalized size of antiderivative = 0.97, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {2823, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int x^3 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right ) \, dx\)

\(\Big \downarrow \) 2823

\(\displaystyle -b n \int \left (-\frac {m x^3}{8}+\frac {1}{4} \log \left (d \left (f x^2+e\right )^m\right ) x^3+\frac {e m x}{4 f}-\frac {e^2 m \log \left (f x^2+e\right )}{4 f^2 x}\right )dx+\frac {1}{4} x^4 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )-\frac {e^2 m \log \left (e+f x^2\right ) \left (a+b \log \left (c x^n\right )\right )}{4 f^2}+\frac {e m x^2 \left (a+b \log \left (c x^n\right )\right )}{4 f}-\frac {1}{8} m x^4 \left (a+b \log \left (c x^n\right )\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{4} x^4 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )-\frac {e^2 m \log \left (e+f x^2\right ) \left (a+b \log \left (c x^n\right )\right )}{4 f^2}+\frac {e m x^2 \left (a+b \log \left (c x^n\right )\right )}{4 f}-\frac {1}{8} m x^4 \left (a+b \log \left (c x^n\right )\right )-b n \left (\frac {1}{16} x^4 \log \left (d \left (e+f x^2\right )^m\right )-\frac {e^2 m \operatorname {PolyLog}\left (2,\frac {f x^2}{e}+1\right )}{8 f^2}-\frac {e^2 m \log \left (e+f x^2\right )}{16 f^2}-\frac {e^2 m \log \left (-\frac {f x^2}{e}\right ) \log \left (e+f x^2\right )}{8 f^2}+\frac {3 e m x^2}{16 f}-\frac {m x^4}{16}\right )\)

Input:

Int[x^3*(a + b*Log[c*x^n])*Log[d*(e + f*x^2)^m],x]
 

Output:

(e*m*x^2*(a + b*Log[c*x^n]))/(4*f) - (m*x^4*(a + b*Log[c*x^n]))/8 - (e^2*m 
*(a + b*Log[c*x^n])*Log[e + f*x^2])/(4*f^2) + (x^4*(a + b*Log[c*x^n])*Log[ 
d*(e + f*x^2)^m])/4 - b*n*((3*e*m*x^2)/(16*f) - (m*x^4)/16 - (e^2*m*Log[e 
+ f*x^2])/(16*f^2) - (e^2*m*Log[-((f*x^2)/e)]*Log[e + f*x^2])/(8*f^2) + (x 
^4*Log[d*(e + f*x^2)^m])/16 - (e^2*m*PolyLog[2, 1 + (f*x^2)/e])/(8*f^2))
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2823

Int[Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))^(r_.)]*((a_.) + Log[(c_.)*(x_)^(n_. 
)]*(b_.))*((g_.)*(x_))^(q_.), x_Symbol] :> With[{u = IntHide[(g*x)^q*Log[d* 
(e + f*x^m)^r], x]}, Simp[(a + b*Log[c*x^n])   u, x] - Simp[b*n   Int[1/x 
 u, x], x]] /; FreeQ[{a, b, c, d, e, f, g, r, m, n, q}, x] && (IntegerQ[(q 
+ 1)/m] || (RationalQ[m] && RationalQ[q])) && NeQ[q, -1]
 

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 213.38 (sec) , antiderivative size = 1031, normalized size of antiderivative = 4.67

method	result	size
risch	\(\text {Expression too large to display}\)	\(1031\)

Input:

int(x^3*(a+b*ln(c*x^n))*ln(d*(f*x^2+e)^m),x,method=_RETURNVERBOSE)
 

Output:

(-1/4*I*Pi*csgn(I*d)*csgn(I*(f*x^2+e)^m)*csgn(I*d*(f*x^2+e)^m)+1/4*I*Pi*cs 
gn(I*d)*csgn(I*d*(f*x^2+e)^m)^2+1/4*I*Pi*csgn(I*(f*x^2+e)^m)*csgn(I*d*(f*x 
^2+e)^m)^2-1/4*I*Pi*csgn(I*d*(f*x^2+e)^m)^3+1/2*ln(d))*(1/4*(I*Pi*b*csgn(I 
*x^n)*csgn(I*c*x^n)^2-I*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-I*Pi*b*cs 
gn(I*c*x^n)^3+I*Pi*b*csgn(I*c*x^n)^2*csgn(I*c)+2*b*ln(c)+2*a)*x^4+1/2*b*x^ 
4*ln(x^n)-1/8*b*n*x^4)-1/8*x^4*a*m-1/4*m/f^2*b*ln(x^n)*e^2*ln(f*x^2+e)-1/4 
*m/f^2*b*n*e^2*dilog((-f*x+(-e*f)^(1/2))/(-e*f)^(1/2))-1/4*m/f^2*b*n*e^2*d 
ilog((f*x+(-e*f)^(1/2))/(-e*f)^(1/2))-1/4*m/f^2*e^2*ln(f*x^2+e)*b*ln(c)+1/ 
16*I*m*x^4*b*Pi*csgn(I*c*x^n)^3+(1/4*b*x^4*ln(x^n)+1/16*x^4*(2*I*b*Pi*csgn 
(I*x^n)*csgn(I*c*x^n)^2-2*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-2*I*b 
*Pi*csgn(I*c*x^n)^3+2*I*b*Pi*csgn(I*c*x^n)^2*csgn(I*c)+4*b*ln(c)-n*b+4*a)) 
*ln((f*x^2+e)^m)-1/8*x^4*ln(c)*b*m-1/8*m*b*ln(x^n)*x^4+1/8*m/f^2*b*n*e^2+1 
/4*m/f*e*x^2*a-1/4*m/f^2*b*n*e^2*ln(x)*ln((-f*x+(-e*f)^(1/2))/(-e*f)^(1/2) 
)-1/4*m/f^2*b*n*e^2*ln(x)*ln((f*x+(-e*f)^(1/2))/(-e*f)^(1/2))-1/16*I*m*x^4 
*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2-1/16*I*m*x^4*b*Pi*csgn(I*c*x^n)^2*csgn(I 
*c)-1/8*I*m/f*e*x^2*b*Pi*csgn(I*c*x^n)^3-3/16*b*e*m*n*x^2/f+1/16*I*m*x^4*b 
*Pi*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)+1/8*I*m/f^2*e^2*ln(f*x^2+e)*b*Pi*c 
sgn(I*c*x^n)^3-1/4*m/f^2*e^2*ln(f*x^2+e)*a+1/8*I*m/f*e*x^2*b*Pi*csgn(I*x^n 
)*csgn(I*c*x^n)^2-1/8*I*m/f*e*x^2*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c) 
+1/16*b*e^2*m*n*ln(f*x^2+e)/f^2+1/4*m/f*b*ln(x^n)*e*x^2+1/8*I*m/f*e*x^2...
 

Fricas [F]

\[ \int x^3 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right ) \, dx=\int { {\left (b \log \left (c x^{n}\right ) + a\right )} x^{3} \log \left ({\left (f x^{2} + e\right )}^{m} d\right ) \,d x } \] Input:

integrate(x^3*(a+b*log(c*x^n))*log(d*(f*x^2+e)^m),x, algorithm="fricas")
 

Output:

integral((b*x^3*log(c*x^n) + a*x^3)*log((f*x^2 + e)^m*d), x)
 

Sympy [F(-1)]

Timed out. \[ \int x^3 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right ) \, dx=\text {Timed out} \] Input:

integrate(x**3*(a+b*ln(c*x**n))*ln(d*(f*x**2+e)**m),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int x^3 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right ) \, dx=\int { {\left (b \log \left (c x^{n}\right ) + a\right )} x^{3} \log \left ({\left (f x^{2} + e\right )}^{m} d\right ) \,d x } \] Input:

integrate(x^3*(a+b*log(c*x^n))*log(d*(f*x^2+e)^m),x, algorithm="maxima")
 

Output:

1/16*(4*b*x^4*log(x^n) - (b*(n - 4*log(c)) - 4*a)*x^4)*log((f*x^2 + e)^m) 
+ integrate(-1/8*((4*(f*m - 2*f*log(d))*a - (f*m*n - 4*(f*m - 2*f*log(d))* 
log(c))*b)*x^5 - 8*(b*e*log(c)*log(d) + a*e*log(d))*x^3 + 4*((f*m - 2*f*lo 
g(d))*b*x^5 - 2*b*e*x^3*log(d))*log(x^n))/(f*x^2 + e), x)
 

Giac [F]

\[ \int x^3 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right ) \, dx=\int { {\left (b \log \left (c x^{n}\right ) + a\right )} x^{3} \log \left ({\left (f x^{2} + e\right )}^{m} d\right ) \,d x } \] Input:

integrate(x^3*(a+b*log(c*x^n))*log(d*(f*x^2+e)^m),x, algorithm="giac")
 

Output:

integrate((b*log(c*x^n) + a)*x^3*log((f*x^2 + e)^m*d), x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

Timed out. \[ \int x^3 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right ) \, dx=\int x^3\,\ln \left (d\,{\left (f\,x^2+e\right )}^m\right )\,\left (a+b\,\ln \left (c\,x^n\right )\right ) \,d x \] Input:

int(x^3*log(d*(e + f*x^2)^m)*(a + b*log(c*x^n)),x)
 

Output:

int(x^3*log(d*(e + f*x^2)^m)*(a + b*log(c*x^n)), x)
 

Reduce [F]

\[ \int x^3 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right ) \, dx=\frac {8 \left (\int \frac {\mathrm {log}\left (x^{n} c \right )}{f \,x^{3}+e x}d x \right ) b \,e^{3} m n +4 \,\mathrm {log}\left (\left (f \,x^{2}+e \right )^{m} d \right ) \mathrm {log}\left (x^{n} c \right ) b \,f^{2} n \,x^{4}-4 \,\mathrm {log}\left (\left (f \,x^{2}+e \right )^{m} d \right ) a \,e^{2} n +4 \,\mathrm {log}\left (\left (f \,x^{2}+e \right )^{m} d \right ) a \,f^{2} n \,x^{4}+\mathrm {log}\left (\left (f \,x^{2}+e \right )^{m} d \right ) b \,e^{2} n^{2}-\mathrm {log}\left (\left (f \,x^{2}+e \right )^{m} d \right ) b \,f^{2} n^{2} x^{4}-4 \mathrm {log}\left (x^{n} c \right )^{2} b \,e^{2} m +4 \,\mathrm {log}\left (x^{n} c \right ) b e f m n \,x^{2}-2 \,\mathrm {log}\left (x^{n} c \right ) b \,f^{2} m n \,x^{4}+4 a e f m n \,x^{2}-2 a \,f^{2} m n \,x^{4}-3 b e f m \,n^{2} x^{2}+b \,f^{2} m \,n^{2} x^{4}}{16 f^{2} n} \] Input:

int(x^3*(a+b*log(c*x^n))*log(d*(f*x^2+e)^m),x)
 

Output:

(8*int(log(x**n*c)/(e*x + f*x**3),x)*b*e**3*m*n + 4*log((e + f*x**2)**m*d) 
*log(x**n*c)*b*f**2*n*x**4 - 4*log((e + f*x**2)**m*d)*a*e**2*n + 4*log((e 
+ f*x**2)**m*d)*a*f**2*n*x**4 + log((e + f*x**2)**m*d)*b*e**2*n**2 - log(( 
e + f*x**2)**m*d)*b*f**2*n**2*x**4 - 4*log(x**n*c)**2*b*e**2*m + 4*log(x** 
n*c)*b*e*f*m*n*x**2 - 2*log(x**n*c)*b*f**2*m*n*x**4 + 4*a*e*f*m*n*x**2 - 2 
*a*f**2*m*n*x**4 - 3*b*e*f*m*n**2*x**2 + b*f**2*m*n**2*x**4)/(16*f**2*n)